91Ó°ÊÓ

• Temperature is$T=273\text{‿é}$
• Pressure is$P=1.00×{10}^{−2}\text{atm}$
• Density is$\mathrm{ÒÏ}=1.24×{10}^{−5}\text{g}/{\text{cm}}^3$

The expression for the RMS speed is given by,

$v=\sqrt{\frac{3RT}{M}}$

Here V is the RMS speed, R is the universal gravitational constant, T is the temperature of the gas, M is the molar mass of the gas.

The relation between mass and molar mass is given by,

$n=\frac{W}{M}$

Here n is the number of moles, w is the mass of the gas,M is the molar mass of the gas.

Since,

$$\begin{gathered} \mathrm{ÒÏ}=\frac{Pm}{RT} \\ RT=\frac{Pm}{\mathrm{ÒÏ}} \end{gathered}$$

Now, rms velocity is given by

$V_{rms}=\sqrt{\frac{3RT}{m}}$

Substitute$\frac{Pm}{\mathrm{ÒÏ}}$ for RT into the above equation,

$\begin{array}{c}{v}_{rms}=\sqrt{\frac{3Pm}{m\mathrm{ÒÏ}}}\\ v_{rms}=\sqrt{\frac{3P}{\mathrm{ÒÏ}}}\end{array}$

Substitute$1.01×{10}^3\text{Pa}$for P and $1.24×{10}^{−2}\text{kg}/{\text{m}}^3$for P into the above equation,

$\begin{array}{c}{V}_{\text{rms}}=\sqrt{\frac{3×1.01×{10}^3}{1.24×{10}^{−2}}}\\ V_{\text{rms}}=494\text{″¾}/\text{s}\end{array}$

Therefore the RMS speed of the gas molecule is$494\text{″¾}/\text{s}$ .

Now we have to find the molar mass.

Since

$\mathrm{ÒÏ}=\frac{Pm}{RT}$

Substitute$1.24×{10}^{−2}\text{kg}/{\text{m}}^3$ for$\mathrm{ÒÏ}$ ,$1.01×{10}^3\text{Pa}$ for P, $8.314\text{ }J/mol\text{ }â‹\dots K$for R ,$273\text{ }K$ for T into the above equation.

$\begin{array}{c}1.24×{10}^{−2}=\frac{1.01×{10}^3×m}{8.31×273}\\ \text{m}=27.9\text{ }g/mol\end{array}$

The molar mass of the gas is around $28\text{ }g/mol$, so the gas molecule is $N_2$.